# numberOfIntegerDigits static method

int numberOfIntegerDigits(
1. dynamic number
)

## Implementation

``````static int numberOfIntegerDigits(dynamic number) {
var simpleNumber = (number.toDouble() as double).abs();
// It's unfortunate that we have to do this, but we get precision errors
// that affect the result if we use logs, e.g. 1000000
if (simpleNumber < 10) return 1;
if (simpleNumber < 100) return 2;
if (simpleNumber < 1000) return 3;
if (simpleNumber < 10000) return 4;
if (simpleNumber < 100000) return 5;
if (simpleNumber < 1000000) return 6;
if (simpleNumber < 10000000) return 7;
if (simpleNumber < 100000000) return 8;
if (simpleNumber < 1000000000) return 9;
if (simpleNumber < 10000000000) return 10;
if (simpleNumber < 100000000000) return 11;
if (simpleNumber < 1000000000000) return 12;
if (simpleNumber < 10000000000000) return 13;
if (simpleNumber < 100000000000000) return 14;
if (simpleNumber < 1000000000000000) return 15;
if (simpleNumber < 10000000000000000) return 16;
// We're past the point where being off by one on the number of digits
// will affect the pattern, so now we can use logs.
return max(1, (log(simpleNumber) / _ln10).ceil());
}``````